Catenyms
| Time Limit: 1000MS | Memory Limit: 65536K | |
| Total Submissions: 8756 | Accepted: 2306 |
Description
A catenym is a pair of words separated by a period
such that the last letter of the first word is the same as the last letter of
the second. For example, the following are catenyms:
A compound catenym is a sequence of three or more words separated by periods such that each adjacent pair of words forms a catenym. For example,
aloha.aloha.arachnid.dog.gopher.rat.tiger
Given a dictionary of lower case words, you are to find a compound catenym that contains each of the words exactly once.
dog.gopher
gopher.rat
rat.tiger
aloha.aloha
arachnid.dog
A compound catenym is a sequence of three or more words separated by periods such that each adjacent pair of words forms a catenym. For example,
aloha.aloha.arachnid.dog.gopher.rat.tiger
Given a dictionary of lower case words, you are to find a compound catenym that contains each of the words exactly once.
Input
The first line of standard input contains t, the
number of test cases. Each test case begins with 3 <= n <= 1000 - the
number of words in the dictionary. n distinct dictionary words follow; each word
is a string of between 1 and 20 lowercase letters on a line by itself.
Output
For each test case, output a line giving the
lexicographically least compound catenym that contains each dictionary word
exactly once. Output "***" if there is no solution.
Sample Input
2 6 aloha arachnid dog gopher rat tiger 3 oak maple elm
Sample Output
aloha.arachnid.dog.gopher.rat.tiger ***
Source
把26个小写字母当成点，每个单词就是一条边。
然后就是求欧拉路径。
为了保证字典序最小，要先排序，加边要按照顺序加。
而且求解的dfs起点要选择下，选择最小的。
1 /* *********************************************** 2 Author :kuangbin 3 Created Time :2014-2-3 13:12:43 4 File Name :E:\2014ACM\专题学习\图论\欧拉路\有向图\POJ2337.cpp 5 ************************************************ */ 6 7 #include <stdio.h> 8 #include <string.h> 9 #include <iostream> 10 #include <algorithm> 11 #include <vector> 12 #include <queue> 13 #include <set> 14 #include <map> 15 #include <string> 16 #include <math.h> 17 #include <stdlib.h> 18 #include <time.h> 19 using namespace std; 20 struct Edge 21 { 22 int to,next; 23 int index; 24 bool flag; 25 }edge[2010]; 26 int head[30],tot; 27 void init() 28 { 29 tot = 0; 30 memset(head,-1,sizeof(head)); 31 } 32 void addedge(int u,int v,int index) 33 { 34 edge[tot].to = v; 35 edge[tot].next = head[u]; 36 edge[tot].index = index; 37 edge[tot].flag = false; 38 head[u] = tot++; 39 } 40 string str[1010]; 41 int in[30],out[30]; 42 int cnt; 43 int ans[1010]; 44 void dfs(int u) 45 { 46 for(int i = head[u] ;i != -1;i = edge[i].next) 47 if(!edge[i].flag) 48 { 49 edge[i].flag = true; 50 dfs(edge[i].to); 51 ans[cnt++] = edge[i].index; 52 } 53 } 54 int main() 55 { 56 //freopen("in.txt","r",stdin); 57 //freopen("out.txt","w",stdout); 58 int T,n; 59 scanf("%d",&T); 60 while(T--) 61 { 62 scanf("%d",&n); 63 for(int i = 0;i < n;i++) 64 cin>>str[i]; 65 sort(str,str+n);//要输出字典序最小的解，先按照字典序排序 66 init(); 67 memset(in,0,sizeof(in)); 68 memset(out,0,sizeof(out)); 69 int start = 100; 70 for(int i = n-1;i >= 0;i--)//字典序大的先加入 71 { 72 int u = str[i][0] - ‘a‘; 73 int v = str[i][str[i].length() - 1] - ‘a‘; 74 addedge(u,v,i); 75 out[u]++; 76 in[v]++; 77 if(u < start)start = u; 78 if(v < start)start = v; 79 } 80 int cc1 = 0, cc2 = 0; 81 for(int i = 0;i < 26;i++) 82 { 83 if(out[i] - in[i] == 1) 84 { 85 cc1++; 86 start = i;//如果有一个出度比入度大1的点，就从这个点出发，否则从最小的点出发 87 } 88 else if(out[i] - in[i] == -1) 89 cc2++; 90 else if(out[i] - in[i] != 0) 91 cc1 = 3; 92 } 93 if(! ( (cc1 == 0 && cc2 == 0) || (cc1 == 1 && cc2 == 1) )) 94 { 95 printf("***\n"); 96 continue; 97 } 98 cnt = 0; 99 dfs(start); 100 if(cnt != n)//判断是否连通 101 { 102 printf("***\n"); 103 continue; 104 } 105 for(int i = cnt-1; i >= 0;i--) 106 { 107 cout<<str[ans[i]]; 108 if(i > 0)printf("."); 109 else printf("\n"); 110 } 111 } 112 return 0; 113 }